smc

Expanding the quadratic and collecting terms gives $y=2x^2-(2a+2b)x+(a^2+b^2)$. For a quadratic of the form $y=A{x}^2+Bx+C$ with $A>0$, $y$ is minimized when $x=-\frac{B}{2A}$, which is the average of the roots. Thus, the quadratic is minimized when $x=\frac{2a+2b}{2\cdot 2}=\frac{a+b}{2}$, which is answer $\boxed{\frac{a+b}{2}}$.